Tag: standardized measurement

Questions Related to standardized measurement

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

Which of the following pairs is not matched?

  1. Coefficient of self-induction : henry

  2. Magnetic flux : weber

  3. Electric flux : volt-meter

  4. Electric capacity : farad-meter

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electric capacitance of a conductor is the ability with which the conductor can hold the charge.
In SI system, unit of capacitance is farad($F$).
Hence, option D is not matched.

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

kilowatt-hour is the unit of ____. 

  1. potential difference

  2. electric power

  3. electric energy

  4. charge

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\text{Kilowatt hour}$ is the $commercial$ unit of $electrical$ ENERGY.

As the power$(means $ $ \dfrac{Work}{time}=\dfrac{energy}{time})$
so $energy=power\times time $

$\text{so unit of energy= unit of power *unit of time =watt *second }$
Kilowatt-hour is just $3600\times 1000$ times  of the above value of energy.

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

The unit of permittivity of free space, $ \varepsilon _ o $ is

  1. $\dfrac {Coulomb} {newton - meter}$
  2. $ \dfrac {newton - meter^2} { Coulomb ^2} $
  3. $\dfrac {Coulomb^2}{newton- meter^2} $
  4. $ \dfrac {Coulomb^2} {(newton-metre)^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From Coulomb's law, F = (1 / (4 pi epsilon_0)) * (q1 * q2 / r^2). Rearranging for epsilon_0 gives units of (Coulomb * Coulomb) / (newton * meter^2), which simplifies to Coulomb^2 / (newton * meter^2).

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

In the eqn. $\left (P+\dfrac {a}{V^2}\right )(V-b)=$ constant, the unit of $a$ is

  1. $dyne\times cm^5$
  2. $dyne\times cm^4$
  3. $dyne/cm^3$
  4. $dyne\times cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Units of both $P $ and $ \dfrac {a}{V^2}$ must be same.
So, $\dfrac {a}{V^2}=P$  $\implies   a=PV^2$ 

Since unit of $P$ is $dyne \ cm^{-2}$ and that of $V$ is $cm^3$.
$\therefore$ Unit of a is  $\dfrac {dyne}{cm^2}(cm^3)^2=dyne\times cm^4$